MATHEMATICAL METHODSFOR PHYSICS

Luca Guido Molinari

Dipartimento di Fisica, Universita degli Studi di Milano,Via Celoria 16, 20133 Milano, Italy.

1

Preface

It is not completely obvious what a course titled “Mathematical Methods forPhysics” should include. Of course, an introduction to complex analysis, Fourierintegral, series expansions ... the list continues but time is limited, and the restis inevitably a matter of choice.In preparing these notes I felt the need to present the selected topics with enoughrigour and amplitude to offer methodological examples, interesting applications.I also grew in awareness of the beauty of the topics, true gems of intellectualachievement, and the giants who made them. Here and there I provide somehystorical notes. The notes are still a “work in progress”, as learning never ends.They cannot replace the vision and depth of books; the good student exploresthe library (and internet) and makes his own discoveries.A textbook with similarities with these notes is: W. Appel, Mathematics forphysics and physicists, Princeton (2007). Of the many topics left out (grouptheory, variational methods, measure and probability theory, differential geom-etry, functional calculus, stochastic methods, advanced linear algebra, ... ),some are really useful; a nice introductory textbook is: M. Stone and P. Gold-bart, Mathematics for Physics, a guided tour for graduate students, CambridgeUniversity Press (2009).

The last line is devoted to my students: it is because of their participationand interest that these notes improved in the years, and are now collected inthis volume. I thank M. Raciti for many useful comments.Milano, 1 january 2014.

Luca Guido Molinari

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INDEX

Part I: COMPLEX ANALYSIS

1. COMPLEX NUMBERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Cubic equation and imaginary numbers. The quartic equation. Beyond the quartic. The

complex field.

2. THE FIELD OF COMPLEX NUMBERS . . . . . . . . . . . . . . . . . . . . . . . . . . 7The field C. Complex conjugation. Modulus. Argument. Exponential. Logarithm. Real

powers. The fundamental theorem of algebra. The cyclotomic equation.

3. THE COMPLEX PLANE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16Straight lines and circles. Simple maps (linear map, inversion, Mobius maps). The stereo-

graphic projection.

4. SEQUENCES AND SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Topology. Sequences (quadratic maps, Julia and Mandelbrot sets). Series (absolute conver-

gence, Cauchy product of series, the geometric series, the exponential series, Riemann’s Zeta

function).

5. COMPLEX FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Continuity. Differentiability and Cauchy-Riemann conditions. Conformal mapping I. Inverse

functions.

6. ELECTROSTATICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41The fundamental solution (the dipole). Simple problems and conformal mapping (thin metal

plate, adjacent thin metal plates). Point charge and semi-infinite conductor (point charge and

conducting half-line, point charge and conducting disk). The planar capacitor (semi-infinite

planar capacitor).

7. COMPLEX INTEGRATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Paths and Curves. Complex integration (two useful inequalities). Primitive. Index of a closed

path.

8. CAUCHY’S THEOREM FOR RECTANGLES . . . . . . . . . . . . . . . . . . . 599. ENTIRE FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Cauchy’s theorem. Cauchy’s integral formula. Other properties. Range of entire functions.

Polynomials

10. CAUCHY’S THEORY FOR HOLOM. FUNCTIONS . . . . . . . . . . 68Cauchy transform.

11. POWER SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Uniform and normal convergence. Power series. The binomial series. Generating functions

ii

and polynomials (Hermite, Laguerre, Chebyshev, Legendre, Hypergeometric series). Differen-

tial equations (Airy’s equation).

12. ANALYTIC CONTINUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Zeros of analytic functions. Analytic continuation. Gamma function (Stirling’s formula,

Digamma function). Analytic maps.

13. LAURENT SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91Laurent series of holomorphic functions. Bessel functions (integer order). Fourier series.

14. THE RESIDUE THEOREM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96Singularities. Residues and their evaluation. Evaluation of integrals (trigonometric integrals,

integrals on the real line, principal value integrals, integrals with branch cut, integrals of hy-

perbolic functions, other examples). Enumeration of zeros and poles. Evaluation of sums.

15. ELLIPTIC FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110Elliptic integrals. Jacobi elliptic functions (derivatives, summation formulae, elliptic integrals,

symmetric integrals). Complex argument. Conformal mapping. General theory. Theta func-

tions.

16. QUATERNIONS AND BEYOND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124Quaternions and vector calculus. Octonions.

Part II: FUNCTIONAL ANALYIS

17. METRIC SPACES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129Metric spaces and completeness. Contractive maps.

18. BANACH SPACES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133Normed and Banach spaces. Maps between normed spaces (the inverse operator). Linear

bounded operators on X (the inverse of a linear operator, the exponential of an operator, the

dual space). The Banach spaces Lp(Ω) (L1(Ω), Riesz-Fisher theorem, Holder and Minkowski

inequalities, Lp(Ω) spaces, L∞(Ω) space).

19. HILBERT SPACES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142Inner product spaces. The Hilbert norm (Bessel and Schwarz’s inequalities, parallelogram

law, polarization formulae). Isomorphism (square summable sequences). Orthogonal systems

(orthogonal polinomials, Legendre polynomials, Hermite polynomials). Linear subspaces and

projections. Complete orthogonal systems. Bargmann’s space.

20. TRIGONOMETRIC SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158Fourier series. Point-wise convergence (Gibbs’ phenomenon, Fourier series on [0, π]). Appli-

cations (Heat equation, Kepler’s equation, Vibrating string, the Euler-Mac Laurin expansion,

Poisson’s summation formula). Fejer sums. Convergence in the mean. Fourier series to Fourier

integrals.

21. BOUNDED LINEAR OPERATORS ON HILBERT SPACES176Linear functionals. Bounded linear operators (integral operators, orthogonal projections, the

position and momentum operators). Unitary operators. Notes on spectral theory.

22. UNITARY GROUPS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .185Stone’s theorem, Weyl operators, rotations (space rotations SO(3), SU(2), unitary represen-

tations).

22. UNBOUNDED LINEAR OPERATORS . . . . . . . . . . . . . . . . . . . . . . . . 194The graph of an operator. Closed operators. The adjoint operator (self-adjointness). Spectral

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theory (the resolvent).

23. SCHWARTZ’S SPACE AND FOURIER TRANSFORM . . . . . 202Introduction, The Schwartz space (seminorms and convergence), The Fourier transform in S .

Convolution product. Applications (the heat equation, Laplace’s equation in the strip).

24. TEMPERED DISTRIBUTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213Introduction. Special distributions (Dirac’s delta, Heaviside’s theta, Principal value, the

Sokhotski-Plemelj formulae). Distributional calculus (derivative, Fourier transform). The

Hilbert transform. Fourier series and distributions. The Kramers Kronig relations. Linear

operators on distributions (generalised eigenvectors, Green functions). The Gel’fand triplet.

25. FOURIER TRANSFORM II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230Fourier transform in L1, Fourier transform in L2 (completeness of the Fourier basis).

26. LAPLACE TRANSFORM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234The Laplace integral. Properties. Inversion (Hankel’s representation of Γ). Convolution.

Mellin transform.

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Part I

COMPLEX ANALYSIS

1

Chapter 1

COMPLEX NUMBERS

1.1 Cubic equation and imaginary numbers.

Imaginary numbers appeared in algebra during the Reinassance, with the solu-tion of the cubic equation12. The problem of solving quadratic equations likex2 + 1 = 0 was considered meaningless, while a real cubic equation always hasa real solution. However, the available method eventually provided the solutionas a sum of terms with imaginary numbers.

The priority for the algebraic solution of the cubic equation is uncertain: itwas probably known to Scipione del Ferro, a professor in Bologna, and NicoloFontana (Tartaglia). The general solution3 was published in the book ArsMagna (1545) by Gerolamo Cardano (Pavia 1501, Rome 1576). It is basedon the algebraic identity

(t− u)3 + 3tu(t− u) = t3 − u3

which he obtained by geometric construction4. After setting x = t − u, theidentity becomes the reduced cubic equation

x3 + 3px+ q = 0 (1.1)

1Before the modern era, the solution was obtained by geometric means; the Persian poetand scientist Omar Khayyam (IX cent.) discussed it as the intersection of a parabola and ahyperbola, by methods that foreran Cartesian geometry.

2Refs: Jacques Sesiano, An Introduction to the History of Algebra, Mathematical World27, AMS; Morris Kline, Mathematical Thought from Ancient to Modern Times, 3 voll, Ox-ford University Press 1972; Carl B. Boyer, A History of Mathematics, Princeton UniversityPress 1985. A source of historical news and pictures is the Mathematics Genealogy Project[www.genealogy.ams.org].

3The use of letters to denote parameters of equations was introduced by Francois Vietefew years later; Cardano solved examples of cubic equations, with all possible signs.

4Consider a cube with edge length t. If three concurring edges are partitioned in segmentsof lengths u and t − u, the cube is cut into two cubes and four parallelepipeds. The totalvolume is t3 = u3 + (t − u)3 + 2tu(t − u) + u2(t − u) + u(t − u)2; simple algebra gives theidentity (W. Dunham, Journey through Genius, the Great Theorems of Mathematics, WileyScience Ed. 1990).

2

with tu = p and t3 − u3 = −q. Therefore, the solution x = t − u of (1.1) isobtained by solving the quadratic equations for t3 and u3, in terms of p and q.Raffaello Bombelli (Bologna 1526, Rome? 1573) in his treatise Algebra was thefirst to regard imaginary numbers as a necessary detour to produce real solutionsfrom real cubic equations. He studied the equation x3 − 15x − 4 = 0, withreal solution x = 4. Cardano’s method works as follows: from tu = −5 andt3−u3 = 4 obtain t6− 4t3+125 = 0 with solutions t3 = 2±

√−121. Imaginary

terms appear:x = (2 +

√−121)1/3 + (2−

√−121)1/3.

Bombelli showed that 2±√−121 = (2±

√−1)3, so that imaginary terms cancel,

and the simple real result x = 4 is recovered.

Exercise 1.1.1. Show that a cubic equation z3 + a1z2 + a2z + a3 = 0 canbe brought to the standard form w3 ± 3w + q = 0 by a linear transformationz = aw + b. For q real, obtain the condition for having three real roots. Findthe three roots through the substitution w = s∓ 1/s.

1.2 The quartic equation.

The Ars Magna also contains the solution of the quartic equation, due to Car-dano’s disciple Ludovico Ferrari (1522, 1565). In modern language, a linearchange of the variable puts it in the form x4 = ax2 + bx+ c. The great idea isthe introduction of an auxiliary parameter y in the equation:

(x2 + y)2 = (a+ 2y)x2 + bx+ (y2 + c).

The parameter is chosen in order to make the right hand side (r.h.s.) of theequation a perfect square of a binomial in x, so that square roots of both sidescan be taken. The condition is the cubic equation for y, 0 = b2−4(a+2y)(y2+c),which may be solved by Cardano’s formula. The value of y is entered in thequartic equation, (x2+y)2 = (a+2y)[x+b/2(a+2y)]2, and a square root bringsit to a couple of quadratic equations in x [Boyer].

Example 1.2.1. To solve the quartic equation x4 − 3x2 − 2x + 5 = 0, rewriteit as (x2 + y)2 = (3+ 2y)x2 +2x− 5+ y2. Choose y such that r.h.s. is a perfectsquare in x, i.e. 2y3 + 3y2 − 10y − 16 = 0 with a solution y = −2. Then theequation is (x2 − 2)2 = −(x − 1)2, i.e. x2 − 2 = ±i(x− 1). The two quadraticequations give the four solutions of the quartic.

The achievement started a great effort to solve higher order equations. Van-dermonde and especially Giuseppe Lagrange (Torino 1736, Paris 1813) empha-sized the role of the permutation group and of symmetric functions. In 1770Lagrange obtained a new method of solution of the quartic equation

z4 + a1z3 + a2z

2 + a3z + a4 = 0.

3

Since it is instructive, we give a brief description of it. The coefficients of theequation are symmetric functions of the roots,

a1 = −∑

i

zi, a2 =∑

i<j

zizj , a3 = −∑

i<j<k

zizjzk, a4 = z1z2z3z4.

Any other symmetric function of the roots is expressible in terms of them.The combination of the roots s1 = z1z2 + z3z4 is not symmetric under all 4!permutations. By doing all permutations only two new combinations appear:s2 = z1z3 + z2z4 and s3 = z1z4 + z2z3.The p = 3 quantities A1 = s1+s2+s3, A2 = s1s2+s1s3+s2s3 and A3 = s1s2s3are symmetric in s1, s2 and s3, and are invariant under permutations of the rootszi. As such, they may be expressed in terms of the coefficients ai: A1 = a2,A2 = a1a3 − 4a4, and A3 = a23 + a21a4 − 4a2a4.The si are the roots of the cubic equation s3 −A1s2 +A2s−A3 = 0, i.e.

s3 − a2s2 + (a1a3 − 4a4)s− a23 + a21a4 − 4a2a4 = 0,

and can be evaluated by Cardano’s method. Once s1, s2 and s3 are obtained,one solves the quadratic equation s1(z1z2) = (z1z2)2 + a4 to get z1z2 and z3z4.In a similar way one gets z1z3, z2z4 and the roots are easily found.

1.3 Beyond the quartic.

For the fifth-order equation Lagrange eagerly tried to guess a polynomial com-bination of the roots that, under the 5! permutations, could produce at mostp = 4 different combinations s1 ... s4 that would solve a quartic equation. Hisdisciple Ruffini (Modena 1765, 1822) showed that such a polynomial should beinvariant under 5!/p permutations of the roots zi, and does not exist.The Norwegian mathematician Niels Henrik Abel (1802, 1829) put the last wordin the memoir On the algebraic resolution of equations published in 1824. Heproved that no rational solution involving radicals and algebraic expressions ofthe coefficients exists for general equations of order higher than four. The iden-tification of the special equations that can be solved by radicals was done byEvariste Galois, by methods of group theory to which he much contributed. Aninteresting result is: an equation is solvable by radicals if and only if, given tworoots, the others depend rationally on them (1830)5.In 1789 E. S. Bring (1786), by exploiting an earlier method by Tschirnhausen(1683), showed that any equation of fifth degree can be brought to the amazinglysimple form z5+q4z+q5 = 0, and then to z5+5z+a = 0 if q4 = 0 (in general, anequation xn+a1xn−1+ · · ·+an = 0 can be reduced to yn+q4yn−4+ · · ·+qn = 0by means of the variable change y = p0+p1x+ · · ·+p4x4 and solving equationsof degree 2 and 3).Charles Hermite succeeded in obtaining a solution of the quintic equation, interms of elliptic functions (1858). Such functions generalize trigonometric ones,

5for a presentation of Galois theory, see V. V. Prasolov, Polynomials, Springer (2004)

4

which are useful to solve the cubic equation6. Soon after Leopold Kroneckerand Francesco Brioschi7 gave alternative derivations8.In 1888 the solution of the general sixth order equation was obtained by Brioschiand Maschke, in terms of hyperelliptic functions.Of course no one would solve even a quartic by the methods described, as effi-cient numerical methods yield the roots with the desired accuracy.

1.4 The complex field

Complex numbers were used in the early XVIII century by Leibnitz, who de-scribed them as entities halfway between existent and nonexistent. Jean Ber-noulli, Abraham De Moivre and, above all, the genius Leonhard Euler (1707,1783) discovered several relations involving trigonometric, exponential and lo-garithmic functions with imaginary argument.The great improvement in the perception of complex numbers as well definedentities was their visualisation as vectors, by the norwegian cartographer CasparWessel (1797) and the mathematician Jean Robert Argand (1806), or as pointsin the plane, by Carl Frederich Gauss. It were Gauss’ authority and investi-gations since 1799, that gave complex numbers a status in analysis. Wessel’swork was recognised only a century later, after its translation in French, andArgand’s paper was much criticized.

In 1833 the Irish mathematician William R. Hamilton (1805, 1865) presentedbefore the Irish Academy an axiomatic setting of the complex field C as a formalalgebra on pairs of real numbers. In 1867 Hankel proved that the algebra ofcomplex numbers is the most general one that fulfils all fundamental laws ofarithmetic (Boyer).

6for example, the equation y3−3y+1 = 0 can be solved with the aid of trigonometric tables:put y = 2 cos x and obtain 0 = 2 cos(3x) + 1; then 3x = 2

3π and 3x = 43π i.e. y1 = 2 cos( 29π)

and y2 = 2 cos( 49π); the other solution is y3 = −y1 − y2.7F. Brioschi (1824, 1897) taught mechanics in Pavia. He then founded Milan’s Politec-

nico (1863), where he taught hydraulics. He participated in Milan’s insurrection, and becamemember of the Parliament. Among his students (in Pavia): Giuseppe Colombo (he inaugu-rated in 1883 in Milan the first thermo-electric generator in continental Europe, by lightingthe lamps of the Scala theatre. The cables were manufactured by the newly born Pirelli.Colombo succeeded to Brioschi in the direction of the Politecnico), Eugenio Beltrami (nonEuclidean geometry, singular values of a matrix, Laplace Beltrami operator in curved space)and Luigi Cremona (painter).

8A discussion of the quintic eq. is in J.V.Armitage and W.F.Eberlein, Elliptic functions,Lon. Math. Soc. Student Text 67 (2006). See also http://wapedia.mobi/en/Bring radical,or V. Barsan, Physical applications of a new method of solving the quintic equation,arXiv:0910.2957v2; G. Zappa, Storia della risoluzione delle equazioni di V e VI grado ...,Rend. Sem. Mat. Fis. Milano (1995).

5

Figure 1.1: Leonhard Euler (1707, 1783) belongs to an impressive geneal-ogy of mathematicians, rooted in Leibnitz and the Bernoullis. Euler spent manyyears in St. Petersburgh, at the dawn of the Russian mathematical school. Hediscovered several important formulae for complex functions, and establishedmuch of the modern notation. His student Joseph Lagrange was the advisor ofFourier and Poisson. Poisson’s students Dirichlet and Liouville mentored illus-trious mathematicians that contributed to the advancement of complex analysisin Paris (on the side of Dirichlet: Darboux, and then Borel, Cartan, Goursat,Picard, and then Hadamard, Julia, Painleve ...; on the side of Liouville: Catalanand then Hermite, Poincare, Pade, Stieltjes ...).

Figure 1.2: Carl Friedrich Gauss (1777, 1855) became a celebrity aftercomputing the orbit of the first asteroid Ceres, discovered and lost of sight bypadre Piazzi in Palermo (1801). To interpolate the best orbit from observedpoints, Gauss devised the Least Squares method. The orbital elements placedCeres in the region were astronomers were searching for the fifth planet, thatfitted in Titius and Bode’s law. Gauss proved the “fundamental theorem ofalgebra”: a polynomial of degree n has n zeros in the complex plane. Heanticipated several results of complex analysis, which he did not publish. Hisgenealogy contains venerable scientists as the astronomers Bessel and Encke, andmathematicians: Dedekind, Sophie Germain, Gudermann, and Georg Riemann.Among Gauss’ “nephews” are Ernst Kummer and Karl Weierstrass.

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Chapter 2

THE FIELD OFCOMPLEX NUMBERS

2.1 The field C

Definition 2.1.1. The field of complex numbers C is the set of pairs of realnumbers z = (x, y) with the binary operations of sum and multiplication

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2), (2.1)

(x1, y1)(x2, y2) = (x1x2 − y1y2, x1y2 + y1x2) (2.2)

The sum is commutative, associative, with neutral element (called zero) (0, 0)and opposite (−x,−y) of (x, y). The multiplication is commutative, associa-tive, with unity (1, 0), inverse element of any nonzero number, and with thedistributive property (z1 + z2)z3 = z1z3 + z2z3.

Exercise 2.1.2. Evaluate the inverse of a non-zero complex number (x, y).

The subset of elements (x, 0) is closed under both operations and identifieswith the field of real numbers R. Thus the field C is an extension of the field R.If an element (x, 0) is identified by x, a pair can be written as

(x, y) = (1, 0)(x, 0) + (0, 1)(y, 0) = x+ iy,

where i is Euler’s symbol for the imaginary unit (0, 1). This is the usual rep-resentation of complex numbers. Additions and multiplications are done byregarding i as a unit with the property i2 = (−1, 0) = −1.

Proposition 2.1.3. C is not an ordered field.

Proof. Suppose that for any pair of different complex numbers it is either z < wor z > w; we show that the rules a < b → a + c < b + c and a < b, c > 0 →ac < bc are violated. Fix i > 0 then i2 > 0 i.e. −1 > 0. Add 1 to get 0 > 1,multiply by i2 > 0 and obtain 0 > −1, which contradicts −1 > 0.

7

2.2 Complex conjugation

The complex conjugate (c.c.) of z = x+ iy is z = x− iy.Properties: z = z (c.c. is an involution), z1 + z2 = z1 + z2 and z1z2 = z1z2.The real numbers x and y are respectively the real and imaginary parts of z:

x = Re z =z + z

2, y = Im z =

z − z

2i.

2.3 Modulus

The modulus of a complex number is |z| =√

x2 + y2. It is |z|2 = zz and|z| = |z|, Re z ≤ |z| and Im z ≤ |z|. The following properties qualify themodulus as a norm and C as a normed space:

|z| ≥ 0, |z| = 0 iff z = 0, (2.3)

|z1z2| = |z1||z2|, (2.4)

|z1 + z2| ≤ |z1|+ |z2| (triangle inequality) (2.5)

Proof. The first is obvious. The second: use |z1z2|2 = z1z2z1z2. Triangleinequality: |z1 + z2|2 = (z1 + z2)(z1 + z2) = |z1|2 + |z2|2 + z1z2 + z1z2; the lasttwo terms are 2Re(z1z2) ≤ 2|z1z2| = 2|z1||z2|. Then |z1 + z2|2 ≤ |z1|2 + |z2|2 +2|z1||z2| = (|z1|+ |z2|)2.

If 1/z is the inverse of z it is |z(1/z)| = 1 i.e. |1/z| = 1/|z|.The modulus simplifies the evaluation of the inverse of a complex number:

1

z=

z

|z|2i.e.

1

x+ iy=

x

x2 + y2− i

y

x2 + y2

Remark 2.3.1. Property (2.4) has two interesting consequences:1) The unit circle |z| = 1 is closed for complex multiplication.2) For any four integers a, b, c, d there are integers p = ad+ bc and q = |ac− bd|such that (a2+b2)(c2+d2) = q2+p2. For example: (1+52)(1+72) = 122+342.

Exercise 2.3.2. Prove that |z1 + z2|2 + |z1 − z2|2 = 2|z1|2 + 2|z2|2 (in a paral-lelogram, the sum of the squares of the diagonals equals the sum of the squaresof the sides).

Exercise 2.3.3. Prove the very useful inequalities:

1√2(|x|+ |y|) ≤ |x+ iy| ≤ |x|+ |y| (2.6)∣

∣

∣|z|− |w|

∣

∣

∣≤ |z − w| (2.7)

Hint: |z − w|2 = (|z|− |w|)2 + 2|z||w|− 2Re(zw).

8

2.4 Argument

A nonzero complex number z = x + iy may be represented as z = |z|(cos θ +i sin θ) (polar representation), where cos θ = x/|z| and sin θ = y/|z|. The numbercos θ + i sin θ belongs to the unit circle, where multiplication acts additively onphases:

(cos θ1 + i sin θ1)(cos θ2 + i sin θ2) = cos(θ1 + θ2) + i sin(θ1 + θ2).

De Moivre’s formula (cos θ + i sin θ)n = cos(nθ) + i sin(nθ) follows.At this stage, Euler’s representation cos θ + i sin θ = eiθ is introduced as adefinition. It is consistent with the power series for cos θ and sin θ, and with therule eiθeiϕ = ei(θ+ϕ). Then we write:

z = |z|(cos θ + i sin θ) = |z|eiθ (2.8)

The phase θ is the argument of z (θ = arg z) and is defined up to integer mul-tiples of 2π. Note the property arg (z1z2) = arg z1 + arg z2.

The principal argument Arg z is the single determination such that

−π < Arg z ≤ π.

The sign of Arg z is the same as Im z. The real negative half-line is a line ofdiscontinuity (branch cut). By definition the points of the real negative axishave Arg z = π.

Arg(ab) = Arg a+Arg b+

⎧

⎪

⎨

⎪

⎩

−2π if π < Arg a+Arg b

0 if −π < Arg a+Arg b ≤ π

2π if Arg a+Arg b ≤ −π(2.9)

Remark 2.4.1. It is Arg(ab) = Arg a + Arg b when −π < Arg a + Arg b ≤ π.The condition corresponds to the geometric fact that the segment with extremaa and b does not cross the branch cut (i.e. the points a and b “see” each other).Proof: If Arg a ≥ 0 then −π < Arg b ≤ π−Arg a i.e. Arg a−π ≤ Arg b < π: allpoints in this angular sector “see” the point a. The same conclusion is found ifArg a < 0.As a Corollary: Arg(a/b) = Arg a−Arg b if the points a and b are in sight.

Other single determinations of the argument are possible, and always havea line of discontinuity from the origin to infinity.For example, if the line is chosen as the imaginary half line z = ix (x ≥ 0), therange of values of this arg is (− 3

2π,12π]. The points on the cut, by definition,

have argument 12π; other values are: arg(1 + i) = π

4 , arg (−1) = −π, arg 1 = 0.

Exercise 2.4.2. Write the numbers 1± i in polar form.

Exercise 2.4.3. Write eia + eib in polar form (a, b ∈ R). Answer: eia + eib =2∣

∣cos a−b2

∣

∣ ei2(a+b)+iω, where ω = 0 or π if cos a−b

2 is positive or negative.

9

Exercise 2.4.4. Show that the numbers

z =1 + is

1− is, s ∈ R

belong to the unit circle. Is the map s → Arg z invertible? If H is a Hermitianmatrix, show that U = (1 + iH)(1− iH)−1 is a unitary matrix.

2.5 Exponential

The exponential of a complex number z = x+ iy is defined as follows:

ez = exeiy = ex(cos y + i sin y) (2.10)

It exists for all z and defines the exponential function onC, with the fundamentalproperty

ezeζ = ez+ζ

The exponential function is periodic, with period 2πi: ez+2πi = ez, for all z.The trigonometric functions

cos z = 12 (e

iz + e−iz), sin z = 12i (e

iz − e−iz)

with (cos z)2 + (sin z)2 = 1, are periodic with period 2π. Note that they areunbounded in the imaginary direction. One also defines tan z = sin z/ cosz andcot z = 1/ tan z. The hyperbolic functions

cosh z = 12 (e

z + e−z), sinh z = 12 (e

z − e−z)

with (cosh z)2−(sinh z)2 = 1, are periodic with period 2πi. They are unboundedin the real direction. One defines tanh z = sinh z/ cosh z and coth z = 1/ tanh z.

Exercise 2.5.1.1) Show that |ez| = ex, ez = ez.2) Find the zeros of sinh(az + b), a, b ∈ R.3) Show that | cosh(x+ iy)| ≤ coshx.

Exercise 2.5.2 (Chebyshev polynomials).Show that cos(nθ) = Tn(t) is a polynomial in t = cos θ of degree n (Chebyshevpolynomial of the first kind). Evaluate the first few polynomials. Show that theyhave the following recursion and orthogonality properties:

Tn+1(t)− 2t Tn(t) + Tn−1(t) = 0 (2.11)

∫ 1

−1

dt√1− t2

Tm(t)Tn(t) =

⎧

⎪

⎨

⎪

⎩

0 n = m

π n = m = 0

π/2 n = m = 0

(2.12)

Prove similar properties for the Chebyshev polynomials of the second kind:

Un(t) =sin[(n+ 1)θ]

sin θ(t = cos θ).

Hint: 2 cos(nθ) = (cos θ + i sin θ)n + c.c.

10

2.6 Logarithm

The logarithm of a complex number z is the exponent that solves the equalityelog z = z. Note that log z is not defined for z = 0. Because of the periodicityof the exponential function, there are an infinite number of solutions: if log z isa solution, also log z + i2kπ is a solution for any integer k. All such solutionsare denoted as log z. The product rule of exponentials implies log(z1z2) =log z1 + log z2. In particular, with z = |z|eiθ one obtains

log z = log |z|+ i arg z + i 2kπ, k ∈ Z (2.13)

While the real part of log z is well determined as the log of a positive realnumber, the imaginary part reflects the same indeterminacy of the argument ofa complex number.It is natural to define the principal logarithm of a number as

Log z = log |z|+ iArg z (2.14)

The Log has a cut of discontinuity along the real negative axis: for vanishingϵ > 0 and x < 0: Log (x + iϵ) = log |x| + iπ (the formula holds also for ϵ = 0,Log (−1) = iπ) and Log (x− iϵ) = log |x|− iπ.

Remark 2.6.1. By Remark 2.4.1, it is Log(ab) = Log a+ Log b when a and bare in sight; it is Log(a/b) = Log a− Log b when a and b are in sight.

In full analogy with the argument, different single-valued determinations ofthe log are possible and always have a line of discontinuity (branch cut) from0 to ∞ (branch points). Therefore, log(−i) is −iπ2 if the principal log (Log) isused. It is i 32π if the log is chosen with cut on the real positive half-line. It is−iπ2 if the cut is on the half-line z = ix, x ≥ 0.

Exercise 2.6.2. Evaluate Log1+i1−i , Log(ie

iθ).

Exercise 2.6.3. What are Log(−z), Log z, Log(1/z), Log(1/z) in terms ofLog z?

Exercise 2.6.4. Prove that, for 0 < θ < π2 : Log (1 − eiθ) = log[2 sin(12θ)] +

i 12 (θ − π).

2.7 Real powers

The real power of a complex number is defined as

za = ea log z (2.15)

In general it is multi-valued (such is the log):

za = |z|a exp (i a argz + i 2πak), k = 0,±1,±2, . . . (2.16)

Let’s consider the various cases:

11

• a ∈ Z: the power z±n is single-valued (z2 = z · z, z−1 = 1/z, etc.).

• a ∈ Q (a = ±p/q, with p and q coprime): the sequence of powers isperiodic, and only q roots are distinct.Example: (1 + i)2/3 = (

√2eiπ/4)2/3 = 21/3 exp(iπ6 + i 43kπ), the values

k = 0, 1, 2 give three distinct powers.

• a is irrational: the set of powers is infinite.Example: (1+i)π = (

√2eiπ/4)π = 2π/2 exp(iπ2/4+i2π2k), k = 0,±1,±2, . . ..

Exercise 2.7.1. Evaluate: 81/3, (−1)1/5, i1/4, (1 − i)1/6.

Exercise 2.7.2. Show that the square roots of z = a+ ib are:

±

⎡

⎣

√√a2 + b2 + a

2+ i

b

2

√

2√a2 + b2 + a

⎤

⎦

Exercise 2.7.3. Show the properties: |za| = |z|a, zazb = za+b, (zζ)a = zaζa

(for multivalued powers, the sets in the two sides must coincide).

Remark 2.7.4. Eq.(2.15) defines in general the power of a complex numberwith complex exponent:

za+ib = e(a+ib) log z

Examples: 1i = ei log 1 = ei(0+2πik)k∈Z = 1, e±2π, e±4π, . . . ;(1 + i)1−i = e(1−i)(log

√2+iπ

4+2πik)k∈Z =

√2eπ(

14+2k)ei(

π4+2πk−log

√2)k∈Z

2.8 The fundamental theorem of algebra.

Among the great advantages of the complex field is the statement that anypolynomial of degree n with complex coefficients has precisely n zeros in C (fun-damental theorem of algebra)1. Therefore, a polynomial always admits theunique factorization:

p(z) = a0(z − z1)(z − z2) · · · (z − zn), (2.17)

where z1, . . . , zn are the zeros (or roots). The proof was provided by theprinceps mathematicorum Carl Friedrich Gauss, in his doctorate dissertation(1799). Through graphical methods he showed that any polynomial p(x+ iy) =u(x, y) + iv(x, y) necessarily has at least one root. The equations u(x, y) = 0and v(x, y) = 0 describe two curves in the plane, and Gauss was able to showthat they have at least an intersection. Before him, Girard (1629), d’Alembert

1This intuitive explanation is from: Timothy Gowers, The Princeton companion to Math-ematics, Princeton University Press, (2009). Let p(z) = zn +a1zn−1 + · · ·+ an, with an = 0.For very large R the set p(Reiθ) ≈ Rneinθ, θ ∈ [0, 2π), approximates a circle of radius Rn

run n times, that contains the origin. For very small R, the set p(Reiθ) ≈ an−1Reiθ + an isa circle does not contain the origin. By continuity, there must be a value R such that the setcontains the origin i.e. an angle θ such that p(Reiθ) = 0.

12

(1748) and Euler (1749), proved the weaker statement that any polynomial withreal coefficients can be factorized into real linear and quadratic terms.

A simple and rigorous proof of the fundamental theorem will be given onthe basis of Liouville’s theorem for entire functions.

2.9 The cyclotomic equation

The roots of the equation zn = 1 are the corners of a regular n-polygon inscribedin the unit circle:

ζk = cos2kπ

n+ i sin

2kπ

n, k = 0, . . . , n− 1.

Since zn − 1 = (z − 1)(zn−1 + · · · + z + 1), the roots ζ1, . . . , ζn−1 solve thecyclotomic equation zn−1 + · · ·+ z + 1 = 0.In 1796 Gauss, then a young student, announced the possibility to construct byruler and compass the regular polygon with n = 17 sides. Five years later hegave a sufficient condition2 for a polygon to be constructible by ruler and com-pass: n = 2k pm1

1 pm2

2 · · · , where the factor 2k accounts for repeated duplicationsof the number of sides of a more basic polygon. The powers pm are either 1 orpowers of a Fermat number (i.e. p = 22

q+1) that is also a prime number3. The

polygons n = 3, 4, 5 (i.e. p = 1, 3, 5) and duplications (n = 6, 8, 10, . . . ) were

known since Euclid’s time. The next Fermat prime number is p = 222

+1 = 17.Gauss was so proud of his discovery, that he asked for a 17-polygon to be carvedon his gravestone4.

Exercise 2.9.1. Let ζkn−1k=0 be the roots of zn = 1. Show that:

ζp0 + · · ·+ ζpn−1 = 0 p = 1, . . . , n− 1 (2.18)

an − bn = (a− b)(a− ζ1b) · · · (a− ζn−1b) (2.19)

Exercise 2.9.2. Solve the equation z5 = 1 for the vertices of a pentagon, andobtain cos 2π

5 = 14 (√5− 1).

Hint: let ζk = ei2πk/5, k = 0, .., 4 be the solutions. Since their sum is zero (thecoefficient of z4 is zero in the equation), also the real part of the sum vanishes:1 + 2 cos 2π

5 + 2 cos 4π5 = 0, from which the solution is found.

Exercise 2.9.3 (Discrete Fourier transform5). Show that the following N ×Nmatrix

Frs =1√N

ei2πN rs

is unitary. Evaluate F 2 and show that F 4 = 1 (Hint: you need the sum ofpowers of the N roots of unity).

2Wantzel (1836) showed that it is also necessary.3Euler showed that the Fermat number 232 + 1 (q = 5) is not a prime number.4The tale of the (28 + 1)−gon is narrated in the nice book Dr. Euler’s fabulous formula

by Paul Nahin, Princeton Univ. Press (2006).5M.L.Mehta, Eigenvalues and eigenvectors of the finite Fourier transform, J. Math. Phys.

28 (1987) 781.

13

Exercise 2.9.4. Consider the following n× n matrix:

S =

⎡

⎢

⎢

⎢

⎢

⎣

0 1. . .

. . .

. . . 11 0

⎤

⎥

⎥

⎥

⎥

⎦

(2.20)

Its action on a vector u ∈ Cn is a cyclic shift of the components: (Su)k = uk+1

and (Su)n = u1.1) Show that Sn−1 = ST (T means transposition) and Sn = 1.2) Find the eigenvalues and the eigenvectors of S.3) Find the spectrum of the periodic Laplacian matrix ∆ = S + ST − 2In

∆ =

⎡

⎢

⎢

⎢

⎢

⎣

−2 1 1

1. . .

. . .. . .

. . . 11 1 −2

⎤

⎥

⎥

⎥

⎥

⎦

4) Find the characteristic poylnomial det(zIn −∆).5) Write down the circulant matrix a0 + a1S + · · · + an−1Sn−1, ai ∈ C, andevaluate its eigenvalues and eigenvectors.

Exercise 2.9.5. Evaluate the characteristic polynomial of the n× n matrix

H0 =

⎡

⎢

⎢

⎢

⎢

⎣

0 1 0

1. . .

. . .. . .

. . . 10 1 0

⎤

⎥

⎥

⎥

⎥

⎦

.

Hint: Let Dk(z) be the determinant of the matrix zIk −H0. Show that it solvesthe recursion Dk+1(z) − zDk(z) + Dk−1(z) = 0. Find the initial conditionsD1(z) and D0(z). Note that:

[

Dk+1(z)Dk(z)

]

= T k

[

D1(z)D0(z)

]

, T =

[

z −11 0

]

By the Cayley-Hamilton theorem, a matrix solves its characteristic equation.The “transfer matrix” T has determinant 1 and trace z, then: T 2− zT +1 = 0.This implies that T k = aT + bI2 with real numbers a, b to be found.

14

Chapter 3

THE COMPLEX PLANE

The modulus of a complex number defines an Euclidean distance between pointsz = (x, y) and z′ = (x′, y′) in the complex plane:

d(z, z′) = |z − z′| =√

(x− x′)2 + (y − y′)2 (3.1)

Results of Cartesian geometry of R2 can be transposed to C. Disks, circles andlines are important in complex analysis; let’s review them in complex notation.

3.1 Straight lines and circles

Given two points a and b in C, the oriented straight line ab has parametricequation z(t) = (1− t)a+ tb, t ∈ R. The restriction 0 ≤ t ≤ 1 traces the closedsegment [a, b] from a to b.A circle C(a, r) with center a and radius r has equation |z − a| = r. Theparameterization

z(θ) = a+ reiθ, 0 ≤ θ < 2π (3.2)

endows the circle with the standard anticlockwise orientation.

Exercise 3.1.1.1) Give conditions for the lines ab and cd to be parallel or orthogonal.2) Find the corners of the squares and of the equilateral triangles having oneside on the segment [a, b].3) Describe the sets: Arg(z − i) = π

3 , |Arg(z − i)| < π3 .

4) Show that the locus |z − a|2 + |z − b|2 = |a − b|2 is a circle with diameter[a, b]. Find the parametric equation of the circle.5) Show that the locus |z − a| = λ|z − b|, λ = 0, is a circle with radius r =

λ|1−λ2| |a − b| and center c = a−λ2b

1−λ2 . The value λ = 1 corresponds to the limit

case of a line (the axis of the segment [a, b]).

15

Exercise 3.1.2.1) Find the equation of the ellipse with foci z1 and z2, major semiaxis length a.2) Study the family of Cassini ovals |z2 − 1| = r2 as r changes. Show that it isa single closed line for r ≥ 1.3) Give the conditions for a point z to be inside the triangle with vertices a, b, c.Answer: z = αa+ βb+ γc, where α+ β + γ = 1 and 0 ≤ α,β, γ ≤ 1.

3.2 Simple maps

In this section we study simple maps w = F (z), where F : C → C. The subjectis of considerable interest and will be fully appreciated after the discussion ofanalytic functions.It is useful to introduce the extended complex plane C by adjoining thepoint ∞ to C with the rules z +∞ = ∞+ z = ∞ and z∞ = ∞z = ∞ (z = 0).Moreover, one puts the conventions z/∞ = 0 (z = ∞), and z/0 = ∞ (z = 0).

3.2.1 The linear map

The linear map w = az + b has one fixed point1 z⋆ = b/(1 − a). By writingw − z⋆ = a(z − z⋆) one obtains

|w − z⋆| = |a||z − z⋆|, arg (w − z⋆) = arg a+ arg (z − z⋆)

Therefore the map is a dilation by a factor |a| of all segments originating fromz⋆ and a rotation of the plane by arg a around the fixed point.Equivalently, the map can be viewed as a rotation by arg a around the originand dilation centred in the origin (z′ = az), followed by a shift (w = z′ + b).

Exercise 3.2.1. Show that a linear map takes circles to circles and straightlines to straight lines.

3.2.2 The inversion map

The map w = 1/z transforms a circle |z| = r centred in the origin into the circle|w| = 1/r. The interior of the unit circle is exchanged with the exterior.By regarding straight lines as circles with a point at infinity, the inversion takescircles to circles (prove it). Then, a straight line that does not contain the originis mapped to a circle through the origin; only straight lines through the origin(containing both 0 and ∞) are mapped to straight lines through the origin.Conversely, circles through the origin are mapped to straight lines, and circlesnot through the origin are mapped to circles.Set w = u+ iv, the image of x+ iy has Cartesian coordinates

u =x

x2 + y2, v = −

y

x2 + y2

1A fixed point of a map is one that is mapped in itself, z⋆ = F (z⋆).

16

x

y

−6

4v=−1/4

u=−1/6u=1/3

v=1/2

Figure 3.1: Inversion map. The circles through the origin of the z plane withcenters on the real or the imaginary axis are pre-images of lines of constant uor v in the w plane. As the lines form an orthogonal grid, also the circles crossat right angles.

A line u(x, y) = U in w−plane is the image of a circle through the origin z = 0with center ( 1

2U , 0). A line v(x, y) = V is the image of a circle again through theorigin with center (0,− 1

2V ). All these circles, that are mapped to the orthogonalgrid of u − v lines, are orthogonal to each other (we’ll give a general proof ofthis).

Exercise 3.2.2. For the inversion map:1) Find the image of the circle |z − i| = 1.2) Show that the circle with center z0 and radius z0 is mapped to the axis of thesegment [0, 1/z0].3) Obtain the image of the line |z − a| = |z − b|, where a, b ∈ C.

3.2.3 Mobius maps

A Mobius map w = M(z) is a linear fractional transformation

M(z) =az + b

cz + d, ad− bc = 0 (3.3)

The map is unchanged if the complex parameters a, b, c, d are multiplied by thesame nonzero value. It has at most two fixed points z⋆ = M(z⋆). For c = 0 theimage of the point −d/c is ∞, and the image of ∞ is a/c.

Proposition 3.2.3. Mobius maps form a group.

Proof. To the Mobius map (3.3) there corresponds an invertible matrix

L =

[

a bc d

]

, detL = 0. (3.4)

17

Such matrices form the linear group of invertible complex 2×2 matricesGL(2,C).IfML(z) is the Mobius map with parameters specified by the matrix L, the com-position of two Mobius maps is the Mobius map ML(ML′(z)) = MLL′(z).

The linear map and the inversion map are Mobius maps with matrices[

a b0 1

]

,

[

0 11 0

]

. (3.5)

If c = 0 the factorization[

a bc d

]

=

[

−ad−bcc

ac

0 1

] [

0 11 0

] [

c d0 1

]

. (3.6)

shows that a Mobius map is a composition of two linear maps with an inversionbetween (the case c = 0 is already a linear map). As a consequence:1) circles are mapped to circles (where a straight line is a circle with point at∞). More precisely, M(z) takes every line and circle passing through −d/c intoa line, and every other line or circle into a circle.2) Mobius maps are bijections from C to C (actually, they are the most generalbijections of the extended complex plane).

A Mobius map can be specified by the requirement that three points (z1, z2, z3)are mapped (in the order) to prescribed points (w1, w2, w3). The choice (0, 1,∞)for the image points gives the Mobius map

M(z) =z − z1z − z3

z2 − z3z2 − z1

. (3.7)

It maps the circle through z1, z2 and z3 to the real axis (the circle that contains0,1 and ∞).

Exercise 3.2.4.1) What are the images of |z − 1| = 1 and |z − 1| = 2 for the map 2 + (1/z)?2) Evaluate the Mobius map that takes (z1, z2, z3) to (w1, w2, w3).3) Identify the Mobius maps that take the upper half-plane H = z : Imz > 0to the unit disk D = w : |w| < 1, and those that take D to D.4) Let 0 and 1 be the fixed points of a Mobius map M(z). Write the generalequation of circles through the two points, and evaluate their images under M(z).5) Evaluate the n-th iterate (M M · · · M)(z) of a Mobius map M(z) (Hint:use the Cayley-Hamilton Theorem for matrices).

3.3 The stereographic projection

The stereographic projection is a bijection among the points of the extendedcomplex plane and the points of the Riemann sphere S2.In a 3D Cartesian frameXY Z consider the spherical surfaceX2+Y 2+(Z− 1

2 )2 =

14 ; it is tangent to the plane Z = 0 in its south pole2. Identify the plane Z = 0

2The Riemann’s sphere is often fixed to have unit radius and center in the origin; otherchoices are possible.

18

with the complex plane z. A segment driven from the north pole (0, 0, 1) to apoint z in the complex plane intersects the spherical surface at the point

X =1

2

z + z

|z|2 + 1, Y =

1

2i

z − z

|z|2 + 1, Z =

|z|2

|z|2 + 1(3.8)

The north pole corresponds to the point ∞ of the extended plane.The Euclidean distance ∥P − P ′∥ in R3 between two points of S2 defines thechordal distance between the two corresponding points in C:

dc(z, z′) =

|z − z′|√

1 + |z|2√

1 + |z′|2(3.9)

The chordal distance differs from the distance established by the modulus,d(z, z′) = |z − z′|. Its limit value 1 is achieved when one of the points is atinfinity.

Exercise 3.3.1. Find the coordinates (X ′, Y ′, Z ′) of the point antipodal to(X,Y, Z) on the Riemann’s sphere. How are their images in C related?(Answer: z′ = −1/z)

Proposition 3.3.2. The stereographic projection maps circles in S2 to circlesin C. The circles through the north pole are mapped to straight lines.

Proof. The locus of points of S2 with Euclidean distance R from a point C ∈ S2

is a circle. The image in C of the circle is the locus dc(z, zC) = R, or |z−zC |2 =R2(1+ |z|2)(1 + |zC |2): the equation of a circle. If the circle in S2 goes throughthe north pole, the image contains z = ∞ and thus it must be R2(1+ |zC|2) = 1to balance infinities, i.e. the equation is linear in z (a straight line).

Proposition 3.3.3. The stereographic projection is conformal (angle-preserving).

Proof. A triangle in C has vertices z, z + ϵ and z + η. The angle in z is givenby Carnot’s formula:

cosα =ϵη + ϵη

2|ϵ||η|The corresponding points on the Riemann’s sphere form a triangle with side-lengths given by the chordal distances. The angle corresponding to α is

cosα′ =dc(z, z + ϵ)2 + dc(z, z + η)2 − dc(z + ϵ, z + η)2

2dc(z, z + ϵ)dc(z, z + η)

=|ϵ|2(1 + |z + η|2) + |η|2(1 + |z + ϵ|2)− |ϵ− η|2(1 + |z|2)

2|ϵ||η|√

(1 + |z + ϵ|2)(1 + |z + η|2)

In the limit |ϵ|, |η| → 0, α′ identifies with the spherical angle and α′ → α.

19

Y

Z

P

N

OH

Q

X

C

Figure 3.2: The stereographic projection maps a point Q ∈ C of coordinate zto a point P = (X,Y, Z) of the Riemann sphere. The correspondence amongcoordinates is obtained through the similitudes: 1 : Z = |z| : (|z|−

√X2 + Y 2)

(ON : HP = OQ : HQ); Re z : X = OQ : OH , Im z : Y = OQ : OH .

Exercise 3.3.4 (Rotations). Show that the Mobius maps that preserve thechordal distance, dc(M(z),M(z′)) = dc(z, z′) for all z, z′ ∈ C, have

|a|2 + |c|2 = |b|2 + |d|2 = 1, ab+ cd = 0, |ad− bc| = 1

They form a group, which is represented by the matrix group U(2) of unitarymatrices on C2. Via the stereographic map, these maps induce rigid transfor-mations of the sphere S2 to itself (rotations and reflections).The subgroup SU(2) corresponds to the choice ad − bc = 1, and induces purerotations.

20

Chapter 4

SEQUENCES ANDSERIES

4.1 Topology

The modulus endows C with the structure of normed space1 and defines a dis-tance between points, d(z, z′) = |z− z′|. Therefore (C, d) is also a metric space.Furthermore, at every point z one may introduce a basis of neighbourhoods,which makes C a topological space. The elements of the basis are the diskscentred in z with radii r > 0:

D(z, r) = ζ : |ζ − z| < r.

The following definitions and statements are important and used thoroughly:

• A set S in C is open if for every point z ∈ S there is a disk D(z, r) wholly inS. The union of any collection of open sets is an open set; the intersectionof two open sets is open.

• A point z ∈ C is an accumulation point of a set S if every disk D(z, r)contains a point in S different from z.

• A boundary point of S is a point z such that every disk D(z, r) containspoints in S and points not in S. The boundary of S is the set ∂S ofboundary points of S.

• A set is closed if it contains all its boundary points. The closure of a setS is the set S = S ∪ ∂S.

• A set S is disconnected if there are two disjoint open sets A and B suchthat S ⊆ A ∪B but S is not a subset of A or B alone. A set is connectedif it is not disconnected.

1Normed, metric and topological spaces are general structures that will be defined later.

21

Proposition 4.1.1. A set S is closed if and only if C/S is open.

Proof. Suppose that S is closed, then for any z /∈ S there is a disk that doesnot contain points in S (otherwise z ∈ ∂S ⊂ S), i.e. C/S is open. On the otherhand, if C/S is open, every point in C/S cannot be in ∂S, i.e. S contains itsfrontier (S is closed).

Definition 4.1.2. A domain is a set both open and connected.

Proposition 4.1.3. Any two points in a domain can be joined by a continuouspolygonal line in the domain.

4.2 Sequences

Complex sequences are maps N → C, and are basic objects in mathematics.They arise in analysis, approximation theory, iteration of maps. Infinite seriesand infinite products are limits of sequences of partial sums and partial products.General statements about sequences are now presented.

Definition 4.2.1. A sequence zn converges to z (zn → z) if |zn − z| → 0 i.e.

∀ϵ ∃Nϵ such that |zn − z| < ϵ, ∀n > Nϵ. (4.1)

Exercise 4.2.2.1) Show that if zn → z and wn → w then: i) zn +wn → z +w, ii) znwn → zw,iii) zn/wn → z/w if wn, w = 0, iv) zn → z.2) Show that a sequence zn is convergent if and only if both Re zn and Im zn areconvergent.3) Show that if zn → z, then |zn| → |z|. Hint: use inequality (2.7).

Definition 4.2.3. A sequence zn is a Cauchy sequence if

∀ϵ > 0 ∃Nϵ such that : |zn − zm| < ϵ, ∀m,n > Nϵ. (4.2)

A convergent sequence is always a Cauchy sequence, but the converse may notbe true. When every Cauchy sequence is convergent to a limit in the space, thespace is termed complete. The Cauchy criterion is then an extremely useful toolto predict convergence, without the need to identify the limit.

Proposition 4.2.4. C is complete

Proof. The inequalities (2.6) imply that xn+ iyn is a Cauchy sequence if andonly if both xn and yn are Cauchy sequences. Since R is complete, theyboth converge. Let x and y be their limits, then |(xn+ iyn)− (x+ iy)| → 0.

22

Figure 4.1: The Mandelbrot set is the locus of c-values such that the sequencezn+1 = z2n + c starting from z0 = 0 remains bounded. The Filled Julia sets forsome values of the parameter c are shown. An initial point z0 picked in a filledJulia generates a bounded sequence.

4.2.1 Quadratic maps, Julia and Mandelbrot sets

The iteration of a map z′ = F (z), with function F : C → C and initial valuez, generates a sequence z, F (z), F (F (z)), ... A sequence depends on the initialvalue, and this dependence may be surprisingly interesting. The problem wasstudied by Pierre Fatou and Gaston Julia in the early 1900. The simplest non-trivial function is

F (z) = z2 + c, c ∈ C.

The quadratic map has two fixed points: z⋆ = z⋆2 + c. Near a fixed point it isF (z) ≈ z⋆+2z⋆(z−z⋆). The number 2|z⋆| depends on c and can be greater, lessor equal to one. The linearized map is accordingly locally expanding, contractingor indifferent, i.e. the distance of images |F (z)− z⋆| is greater, less or equal tothe distance |z − z⋆|.The sequences that escape to ∞ define a set of initial points Ac(∞) called

the attraction basin of ∞. Of course, if z ∈ Ac(∞) also F (z) and the wholesequence belongs to it. It was proven that it is an open and connected set.The complementary set Kc is the filled Julia set. It contains the initial pointsof bounded sequences and it is a closed and bounded (i.e. compact) set. Theboundary is the Julia set Jc. Each set Ac(∞), Kc and Jc is invariant under the

23

map.2 Fatou and Julia proved the theorem: If the sequence 0, c, F (c), F (F (c)), ...diverges to∞, i.e. 0 ∈ Ac(∞), then Jc is totally disconnected, whereas if 0 ∈ Kc,then Jc is connected.While at IBM, in 1980, Benoit Mandelbrot (1924-2010) studied with the aid ofa computer the properties of invariant Julia sets of the quadratic map and morecomplex ones, and disclosed the beauty of their fractal structures (see the bookby Falconer for further reading, and Wikipedia for wonderful pictures). TheMandelbrot set (1980) is the set of parameters c ∈ C such that Jc is connected.Again, it is a wonderful fractal.

Exercise 4.2.5. Consider the linear map zn+1 = azn + b. Write the generalexpression of zn in terms of z0. When is the sequence bounded?

Exercise 4.2.6. Study the fixed points z⋆ of the exponential map z′ = ez.Show that they come in pairs x± iy with x > 0, |y| > 1 (then the linearized mapz′ = z⋆ + z⋆(z − z⋆) is locally expanding)3.

4.3 Series

Infinite sums were studied long before sequences. The oldest known series ex-pansions were obtained by Madhava (1350, 1425), the founder of the Keralaschool of astronomy and mathematics. He developed series for trigonometricfunctions, including an error term. The one for arctan x enabled him to eval-uate π up to 11 digits. His work may have influenced European mathematicsthrough transmission by the Jesuits. The same series were rediscovered by Gre-gory two centuries later.Oresme in the XIV century, Jakob & Johann Bernoulli (Tractatus de seriebusinfinitis, 1689), and Pietro Mengoli (1625, 1686), discovered and rediscoveredthe divergence of the Harmonic series 1+ 1

2 +13 + . . . In the Tractatus, the con-

vergence of∑

k 1/k2 was also proven, but the sum was evaluated later (1734) by

Johann’s prodigious student Leonhard Euler. Euler also proved that the sumof the reciprocals of prime numbers is divergent4.Christian Huygens asked his student Leibnitz to evaluate the sum of reciprocalsof triangular numbers5. The result (the sum is 2) was obtained after notingthat 1

k(k+1) = 1k − 1

k+1 (the sum is a telescopic series).

2For c = 0 the sets are identified easily. Clearly A0(∞) is the set |z| > 1, K0 is the set|z| ≤ 1 (the filled Julia set) and J0 (the Julia set) is the unit circle |z| = 1. The action ofz → z2 on the points ei2πθ ∈ J0 (θ ∈ [0, 1]), is the map θn+1 = 2θn (mod 1). A point θ0after k iterations of the map is again θ0 if (2k − 1)θ0 is an integer. Then the initial point θ0generates a periodic sequence (periodic orbit of the map) of period k. Only rational angles giverise to periodic orbits. The irrational ones spread on the unit circle and the map is chaotic.

3The map is chaotic, and is studied in arXiv:1408.1129.4For nice accounts see P. Pollack, Euler and the partial sums of the prime harmonic series,

http://alpha.math.uga.edu/ pollack/eulerprime.pdf; M.Villarino, Mertens Proof of MertensTheorem, arXiv:math/0504289.

5The triangular numbers are n1 = 1, nk+1 = nk + k = 1 + 2 + · · ·+ k = k(k+1)2 .

24

Series gained rigour after Cauchy, who defined convergence in terms of the se-quence of partial sums.

Given a sequence of complex numbers ak one constructs the partial sumsAn =

∑nk=0 ak. If the sequence An converges to a finite limit A, the limit is the

sum of the series

A =∞∑

k=0

ak.

If∑∞

k ak = A and∑∞

k bk = B, where A and B are finite, the series∑∞

k (ak±bk)is convergent and the sum is A±B.

Proposition 4.3.1. A necessary and sufficient condition for a series to con-verge is that the sequence of partial sums is a Cauchy sequence:

∀ϵ > 0 ∃Nϵ such that ∀m > Nϵ and ∀n > 0 :

∣

∣

∣

∣

∣

m+n∑

k=m+1

ak

∣

∣

∣

∣

∣

< ϵ. (4.3)

In particular for n = 1 it is |am+1| < ϵ. Therefore a necessary condition forconvergence is |ak| → 0.

4.3.1 Absolute convergence

The inequality |∑m+n

m+1 ak| ≤∑m+n

m+1 |ak| implies that if∑∞

k=1 |ak| converges,then also

∑∞k=1 ak does.

Definition 4.3.2. A series∑

k ak is absolutely convergent if∑

k |ak| is conver-gent.

Absolute convergence is a sufficient criterion for convergence; moreover it dealswith series in R. We state but not prove the important property

Proposition 4.3.3. The sum of an absolutely convergent series does not changeif the terms of the series are permuted 6:

∑

k ak =∑

k aπ(k).

The following are sufficient conditions for absolute convergence of the series∑

k ak. They must hold for k greater than some N :

• Comparison (Gauss):

|ak| < bk, where∑

k

bk < ∞

6A convergent series that is not absolutely convergent is conditionally convergent. Theseries

!∞k=1 i

k/k is conditionally convergent as it is the sum of two convergent series!

k≥1(−1)k/(2k) + i!

k≥0(−1)k/(2k + 1). Riemann proved the surprising result that byrearranging terms of a conditionally convergent real series one can obtain any limit sum in R.

25

• Ratio (D’Alembert): for ak = 0

lim supk

|ak+1||ak|

< 1

• Root (Cauchy-Hadamard):

lim supk

k√

|ak| < 1

Proposition 4.3.4 (Cesaro). If the sequences |ak|1/k and |ak+1||ak| converge to

finite limits, the limits are equal. The limit coincides with lim sup.

4.3.2 Cauchy product of series

The Cauchy product of two series is the series of finite sums:[ ∞∑

k=0

ak

][ ∞∑

r=0

br

]

=∞∑

k=0

(

k∑

r=0

arbk−r

)

(4.4)

Proposition 4.3.5 (Franz Mertens, 18747). If a series is absolutely convergentto A and another is convergent to B, their Cauchy product is convergent to AB.

Proposition 4.3.6. If two series are absolutely convergent to A and B, theirCauchy product is absolutely convergent to AB.

Proof.

n∑

k=0

|ck| =n∑

k=0

∣

∣

∣

∣

∣

k∑

r=0

arbk−r

∣

∣

∣

∣

∣

≤n∑

k=0

k∑

r=0

|ar||bk−r|

=n∑

r=0

|ar|n∑

k=r

|bk−r| =n∑

r=0

|ar|n−r∑

k=0

|bk| ≤ AB

Since the sequence∑n

k=0 |ck| is non decreasing and bounded above, it is conver-gent. Then Ck =

∑nk=0 ck is absolutely convergent to a limit C. The limit does

not depend on rearrangements of terms and is the sum of all possible productsarbs. It can be evaluated as follows: C = limk(AkBk) = (limk Ak)(limk Bk) =AB.

4.3.3 The geometric series

The partial sums Sn = 1 + z + . . . + zn can be evaluated from the identitiesSn+1 = Sn + zn+1 and Sn+1 = 1 + zSn:

n∑

k=0

zk =1− zn+1

1− z, z = 1 (4.5)

7see Boas, http://www.math.tamu.edu/ boas/courses/617-2006c/sept14.pdf

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If |z| < 1, it is limn→∞ zn = 0 and Sn(z) converges to the simple but funda-mental geometric series

∞∑

n=0

zn =1

1− z, |z| < 1 (4.6)

The geometric series is useful for assessing convergence of series by the com-parison test. Consider for example the Jacobi theta function (for simplicity, we

restrict to x ∈ R): ϑ3(x, q) = 1 + 2∑∞

m=1 qm2

cos(2mx), |q| < 1. The series isabsolutely convergent:

|ϑ3(x, q)| ≤ 1 + 2∞∑

m=1

|q|m2

≤ 1 + 2∞∑

m=1

|q|m =1 + |q|1− |q|

Exercise 4.3.7. Evaluate the sums: 1 + 2 cosx + · · · + 2 cosnx and sinx +sin 2x+ · · ·+ sinnx.

Exercise 4.3.8. For a > 0, b real, obtain the sums:

∞∑

k=0

e−ka cos kb =1

2+

sinh a

2 cosha− 2 cos b,

∞∑

k=0

e−ka sin kb =sin b

2 cosha− 2 cos b

Hint: evaluate the partial sums∑

k e−k(a+ib) and separate Im and Re parts.

Exercise 4.3.9. Write (1 − z2)−1 as a geometric series in z2, as the productof two geometric series, as a linear combination of two geometric series.

4.3.4 The exponential series

The sequence of partial sums En(z) = 1+z+ · · ·+ 1n!z

n is absolutely convergentfor any z (ratio test):

∣

∣

∣

∣

zn+1

(n+ 1)!

∣

∣

∣

∣

∣

∣

∣

∣

zn

n!

∣

∣

∣

∣

−1

=|z|

n+ 1< 1

for n large enough. The limit of En(z) is the exponential function:

ez =∞∑

n=0

zn

n!, z ∈ C (4.7)

The Cauchy product of two exp series is an exponential series: ez+ζ = ezeζ .It is curious to observe that although ez can never be zero, its polynomialapproximations En have an increasing number n of complex zeros that fly toinfinity8.

Exercise 4.3.10. Show that, according to (4.7), ex+iy = ex(cos y + i sin y).

8Szego (1924) proved that in the large n limit the zeros of En(z), divided by n, distributeon the curve |ze1−z| = 1 and |z| ≤ 1.

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4.3.5 Riemann’s Zeta function

The following series is of greatest importance in number theory9, and is oftenencountered in physics:

ζ(z) =∞∑

n=0

1

(n+ 1)z, Re z > 1 (4.8)

Since |nz | = |ez logn| = nRez, the series converges absolutely for Re z > 1. Thevalues of the function are known at even integers; the following ones are veryuseful (they result, for example, from certain Fourier series, see example 20.2.5):

ζ(2) =π2

6, ζ(4) =

π4

90

Exercise 4.3.11. Prove in the order:

∞∑

n=0

1

(2n+ 1)z= (1 − 2−z)ζ(z),

∞∑

n=0

(−1)n

(n+ 1)z= (1− 21−z)ζ(z) (4.9)

(since Riemann’s series is absolutely convergent, the sum on even and odd nmay be evaluated separately).By collecting in the first series the terms 2n+ 1 that are multiples of 3, obtainthe sum on integers that are not divided by 2 and 3:

∑

n=2k,3k

1

nz=

(

1−1

2z

)(

1−1

3z

)

ζ(z)

The process outlined is iterated and gives the famous representation of Rie-mann’s Zeta function as an infinite product on prime numbers larger than 1:

1

ζ(z)=∏

p

(

1−1

pz

)

(4.10)

Exercise 4.3.12. Show that 10

∫ 1

0dx

∫ 1

0dy

1

1− xy= ζ(2)

(hint: expand the fraction in geometric series)

9H. M. Edwards, Riemann’s Zeta Function, Dover.10This type of integral representation was used to prove irrationality of ζ(2) and ζ(3) in a

way simpler than Apery’s proof of 1978 (see arXiv:1308.2720).

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